# Express the inradius of a right triangle

Imagine the right-angled $\triangle ABC$ with an right angle at $C$ and side-lengths $|AB|= c$, $|BC|=a$, $|CA|=b$. Let $r$ be the radius of the inscribed circle. Then it follows that...

• (a) $r = \frac12(a + b − c)$
• (b) $r = \frac12(c − a − b)$
• (c) $r = \frac12(3a + 2b − 2c)$
• (d) $r = \frac12(2c − a − b)$

Why is alternative (a) correct? I can yet not fathom the reasoning behind it.

• See this answer to the question "Radii of inscribed and circumscribed circles in right-angled triangle", noting that $c$ in this question is $2R$ in that one. (Should this question be marked as a duplicate of the other?) – Blue Apr 16 '18 at 20:56
• I can solve for r when I do get to (b-r)+(a-r)=c but it is not clear to me why that equal sign is correct. How would I know that? – IGotAQuestion Apr 16 '18 at 21:17
• Since my proof-without-words is insufficient, I'll let someone else answer that. – Blue Apr 16 '18 at 21:19
• Would you recommend me to re-post my question in order to increase my chances of getting an answer? I would not mind if you deleted this post so that I could repost it. I am sincerely and desperately looking for an answer. Thanks in advance. – IGotAQuestion Apr 17 '18 at 10:49
• Unfortunately, not every question gets attention (there are zillions posted every day). Reposting is strongly discouraged. The official way to "bump" your question is to edit it with more detail. In this case, you might tell that you've read all the answers (not just mine) to the question I mentioned before but found them lacking; explain, to the best of your ability, what it is that you don't understand about those solutions. This will help potential answerers address your concerns without repeating approaches that you find unhelpful. Good luck! – Blue Apr 17 '18 at 11:52